Question: Roselyn is driving to visit her family, which lives $150$ kilometers away. Her average speed is $60$ kilometers per hour. The car's tank has $20$ liters of fuel at the beginning of the drive, and its fuel efficiency is $6$ kilometers per liter. Fuel costs $0.60$ dollars per liter. What is the price for the amount of fuel that Roselyn will use for the entire drive?
There can be many ways to solve this problem. Here, we will do this by thinking about units. Let's say it will cost Roselyn $x\,\text{dollars}$ to drive $150\,\text{kilometers}$. How can we relate these two quantities with an equation? $\begin{aligned} 150\,\text{kilometers}\cdot y\,\dfrac{\text{dollars}}{\text{kilometer}}=x\,\text{dollars} \end{aligned}$ So in order to find the cost $x$, we need to figure out the value of $y$, which is the rate of the price per kilometer. Notice what other information we are given: $60\,\dfrac{\text{kilometers}}{\text{hour}}$ $20\,\text{liters}$ $6\,\dfrac{\text{kilometers}}{\text{liter}}$ $0.6\,\dfrac{\text{dollars}}{\text{liter}}$ Which of these quantities can help us calculate a rate whose units are $\dfrac{\text{dollars}}{\text{kilometer}}$ ? We can combine the following quantities: $\begin{aligned} &\phantom{=}\dfrac{0.6\,\dfrac{\text{dollars}}{\text{liter}}}{6\,\dfrac{\text{kilometers}}{\text{liter}}} \\\\ &=\dfrac{0.6}{6}\,\dfrac{\text{dollars}}{\cancel\text{liter}}\cdot\dfrac{\cancel\text{liters}}{\text{kilometer}} \\\\ &=0.1\,\dfrac{\text{dollars}}{\text{kilometer}} \end{aligned}$ Now we can plug that in the original equation: $\begin{aligned} 150\,\cancel\text{kilometers}\cdot 0.1\,\dfrac{\text{dollars}}{\cancel\text{kilometer}}&=x\,\text{dollars} \\\\ 15\,\text{dollars}&=x\,\text{dollars} \end{aligned}$ In conclusion, the price for the amount of fuel that Roselyn will use for the entire drive is $15$ dollars.